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CAT 2025Slot 1QAQues & Sol

AlgebraHard

Question

The number of non-negative integer values of k for which the quadratic equation $x^{2}-5x+k=0$ has only integer roots, is

Solution

1. Concept Used

  • Topic: Quadratic Equations — Discriminant and Integer Roots
  • Formula: For a quadratic $ax^2 + bx + c = 0$, roots are integers only when the discriminant $D = b^2 - 4ac$ is a perfect square (including zero), and the resulting root expression yields an integer.

2. Calculation

The given equation is $x^2 - 5x + k = 0$.

For this equation to have integer roots, two conditions must hold simultaneously:

  1. The discriminant $D = 25 - 4k$ must be a perfect square (i.e., $D = m^2$ for some non-negative integer $m$).
  2. The roots $x = \frac{5 \pm m}{2}$ must be integers, which requires $5 \pm m$ to be even, meaning $m$ must be odd.

Also, since $k \geq 0$ (non-negative), we need $D = 25 - 4k \geq 0$, giving $k \leq 6.25$, so $k \in {0, 1, 2, 3, 4, 5, 6}$.

Now we check which values of $k$ make $D$ a perfect square with $m$ odd:

Case $k = 0$: $D = 25 - 0 = 25 = 5^2$. Here $m = 5$ (odd) ✅. Roots: $x = \frac{5 \pm 5}{2} = 5$ or $0$. Both integers. ✅

Case $k = 1$: $D = 25 - 4 = 21$. Not a perfect square. ❌

Case $k = 2$: $D = 25 - 8 = 17$. Not a perfect square. ❌

Case $k = 3$: $D = 25 - 12 = 13$. Not a perfect square. ❌

Case $k = 4$: $D = 25 - 16 = 9 = 3^2$. Here $m = 3$ (odd) ✅. Roots: $x = \frac{5 \pm 3}{2} = 4$ or $1$. Both integers. ✅

Case $k = 5$: $D = 25 - 20 = 5$. Not a perfect square. ❌

Case $k = 6$: $D = 25 - 24 = 1 = 1^2$. Here $m = 1$ (odd) ✅. Roots: $x = \frac{5 \pm 1}{2} = 3$ or $2$. Both integers. ✅

The valid values are $k = 0, 4, 6$, giving us 3 values in total.

3. Solution

Answer = 3 ✅

There are exactly 3 non-negative integer values of $k$ (namely $k = 0, 4, 6$) for which the quadratic equation $x^2 - 5x + k = 0$ has only integer roots.